Complete and balance the following equations, and identify the oxidizing and reducing agents: (a) Cr 2 O 7 2- (aq) + I - (aq) → Cr 3+ (aq) + IO 3 - (aq) (acidic solution) (b) MnO 4 - (aq) + CH 3 O(1aq) → Mn 2+ (aq) + HCOOH(aq) (acidic solution) (c) I 2 (s) + OCl - (aq) → IO 3 - (aq) + Cl - (aq) (acidic solution)

Hi everyone here we have a question asking us to balance the following redox reaction under acidic conditions and identify the oxidizing and reducing agents. We have chlor orrick acid plus hydrazine forms gold plus nitrogen plus hydrochloric acid. So our first step is to separate the whole reaction into half reactions. So we're going to have chlor oric acid forms gold plus hydrochloric acid and we're going to have hydrazine forms nitrogen. Next we're going to balance the non hydrogen and non oxygen elements. So we need to balance our chlorine on our first reaction. So we'll have chlor oric acid forms gold Plus four hydrochloric acid and that's balancing our chlorine. And in the next one our nitrogen is already balanced. So it's going to stay the same Now we need to balance oxygen by adding water to the sides that need it. So our first one is good and our second one is good. So we don't need to do anything there. And next we're going to balance our hydrogen by adding hydrogen to the sides that need it. So we'll have chlor orrick acid plus three hydrogen forms gold Plus four hydrochloric acid. And for our next reaction we'll have hydrazine forms nitrogen plus four hydrogen and now we need to balance the charge is we're going to add electrons to the more positive sides. So we're going to have cooler Orrick acid plus three hydrogen plus three electrons forms gold Plus four hydrochloric acid. And for our next one we're going to have hydrazine forms nitrogen plus four hydrogen plus four electrons. And now we need to balance the electrons on the two half reactions. So we're going to multiply this one by four. We're going to multiply this one x 3. So we are going to end up with four core orrick acid, plus 12 hydrogen plus 12 electrons, forms four gold plus 16 hydrochloric acid. And we're going to have three. Hi Justine forms three nitrogen plus 12 hydrogen plus 12 electrons. And now we need to add the two reactions. So our 12 hydrogen and our total electrons are going to cancel out And that's going to give us a total of four chlor oric assets. Plus three hydrazine forms four gold Plus 16 hydrochloric acids plus three nitrogen and losing electrons is oxidation. So our reducing agent is hydrazine gaining electrons, is reduction and are oxidizing agents. Are oxidizing agent is chlor orrick acid and those are our final answers. Thank you for watching. Bye.

Ch.20 - Electrochemistry

Brown - Chemistry: The Central Science 15th Edition

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Brown 15th Edition

Ch.20 - Electrochemistry

Problem 27a,b,c

Chapter 20, Problem 27a,b,c

Complete and balance the following equations, and identify the oxidizing and reducing agents: (a) Cr₂O₇^2-(aq) + I^-(aq) → Cr³⁺(aq) + IO₃^-(aq) (acidic solution) (b) MnO₄^-(aq) + CH₃O(1aq) → Mn²⁺(aq) + HCOOH(aq) (acidic solution) (c) I₂(s) + OCl^-(aq) → IO₃^-(aq) + Cl^-(aq) (acidic solution)

Verified step by step guidance

Identify the oxidation states of each element in the reactants and products to determine which elements are oxidized and reduced.

Write separate half-reactions for the oxidation and reduction processes. For this reaction, identify the half-reaction where iodine (I) is oxidized to iodate (IO_3^-) and the half-reaction where dichromate (Cr_2O_7^{2-}) is reduced to chromium (Cr^{3+}).

Balance each half-reaction for mass and charge. Start by balancing the atoms other than O and H, then balance O atoms by adding H_2O, and H atoms by adding H^+. Finally, balance the charges by adding electrons (e^-).

Multiply each half-reaction by an appropriate factor so that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.

Add the balanced half-reactions together, canceling out the electrons and any other species that appear on both sides, to obtain the balanced overall equation. Identify the oxidizing agent (the species that is reduced) and the reducing agent (the species that is oxidized).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Redox Reactions

Redox reactions, or oxidation-reduction reactions, involve the transfer of electrons between species. In these reactions, one substance is oxidized (loses electrons) while another is reduced (gains electrons). Understanding the changes in oxidation states of the elements involved is crucial for identifying the oxidizing and reducing agents.

Oxidizing and Reducing Agents

An oxidizing agent is a substance that gains electrons and is reduced in a chemical reaction, while a reducing agent is one that loses electrons and is oxidized. Identifying these agents is essential for understanding the dynamics of redox reactions, as they dictate the flow of electrons and the overall reaction mechanism.

Balancing Redox Reactions

Balancing redox reactions involves ensuring that the number of atoms and the charge are equal on both sides of the equation. This often requires the use of half-reaction methods, where the oxidation and reduction processes are balanced separately before combining them. In acidic solutions, the addition of H+ ions and water may be necessary to achieve balance.

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Balancing Basic Redox Reactions

Related Practice

Textbook Question

Complete and balance the following half-reactions in basic solution. In each case, indicate whether the half-reaction is an oxidation or a reduction.

a. O2(𝑔)⟶H2O(𝑙)

b. Mn2+(𝑎𝑞)⟶MnO2(𝑠)

c. Cr(OH)3(𝑠)⟶CrO42−(𝑎𝑞)

d. N2H4(𝑎𝑞)⟶N2(𝑔)

Textbook Question

Complete and balance the following half-reactions in basic solution. In each case, indicate whether the half-reaction is an oxidation or a reduction. c. Cr(OH)3(𝑠)⟶CrO42−(𝑎𝑞)

Textbook Question

Complete and balance the following half-reactions in acidic solution. In each case indicate whether the half-reaction is an oxidation or a reduction. b. H2SO3(𝑎𝑞)⟶SO42−(𝑎𝑞)

Textbook Question

Complete and balance the following equations, and identify the oxidizing and reducing agents: MnO₄^-(aq) + Br^-(aq) → MnO₂(s) + BrO₃^-(aq) (basic solution)

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Textbook Question

Complete and balance the following equations, and identify the oxidizing and reducing agents:

a. MnO4−(𝑎𝑞)+CH3OH(𝑎𝑞)⟶Mn2+(𝑎𝑞)+HCOOH(𝑎𝑞)(acidic solution)

b. As2O3(𝑠)+NO3−(𝑎𝑞)⟶H3AsO4(𝑎𝑞)+N2O3(𝑎𝑞)(acidic solution)

c. Pb(OH)42−(𝑎𝑞)+ClO−(𝑎𝑞)⟶PbO2(𝑠)+Cl−(𝑎𝑞)(basic solution)

Textbook Question

Complete and balance the following equations, and identify the oxidizing and reducing agents: As₂O₃(s) + NO₃^-(aq) → H₃AsO₄(aq) + N₂O₃(aq) (acidic solution)

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Complete and balance the following equations, and identify the oxidizing and reducing agents: (a) Cr2O72-(aq) + I-(aq) → Cr3+(aq) + IO3-(aq) (acidic solution) (b) MnO4-(aq) + CH3O(1aq) → Mn2+(aq) + HCOOH(aq) (acidic solution) (c) I2(s) + OCl-(aq) → IO3-(aq) + Cl-(aq) (acidic solution)

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Key Concepts

Redox Reactions

Oxidizing and Reducing Agents

Balancing Redox Reactions